# Probability of getting at least one ace

You have a normal 52 card deck. Note that your hand does not depend on order. You are dealt two cards. What is the probability you will get at least one ace?

How do I approach this problem?

• What's the probability of getting no aces? – Angina Seng Apr 27 '17 at 4:23
• @LordSharktheUnknown remaining 48 cards are not aces – socrates Apr 27 '17 at 4:26
• Is it $1-(48/52C2)$? – socrates Apr 27 '17 at 4:27
• How many hands are possible? How many hands with no aces are possible? – shardulc says Reinstate Monica Apr 27 '17 at 4:27
• @shardulc $52C2$; $52C48$ – socrates Apr 27 '17 at 4:29

This you do by saying that there are $48$ non-aces in the deck, so you pick any two of those in $\binom{48}2$ ways, and the ways of picking two cards from the complete deck is $\binom{52}2$. So the probability of not picking an ace, is $\frac{\binom{48}2}{\binom{52}2}$.
The answer you are then looking for is $1 - \frac{\binom{48}2}{\binom{52}2}$.
EDIT : For those who want a nice looking answer,it's $\frac{33}{221}$.
The denominator is the number of 2 card hands, which is ${52 \choose 2}$. The numerator is the number of 2 card hands which contain at least one ace. This quantity is ${52 \choose 2}$ minus the number of 2 card hands which do not contain any aces, and hence is equal to ${52 \choose 2} - {48 \choose 2}$. Thus, the probability in question is $\frac{{52 \choose 2}-{48 \choose 2}}{{52 \choose 2}}$.
Let $A$ be the event of getting at least one ace, thus $A'$ be defined as the complement event which is getting no ace. $P(A) = 1 - P(A') = 1 - \dfrac{\binom{48}{2}}{\binom{52}{2}}=...$